The figure shows a liquid of given density flowing steadily in a horizontal tube of varying cross-section. Cross-sectional areas at A is , and B is . If the speed of liquid at B is , then is [2023]

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Solution

Q.
The figure shows a liquid of given density flowing steadily in a horizontal tube of varying cross-section. Cross-sectional areas at A is $1.5 {cm}^{2}$ , and B is $25 {mm}^{2}$ . If the speed of liquid at B is $60 cm/s$ , then $(P_{A} - P_{B})$ is

(Given $P_{A}$ and $P_{B}$ are liquid pressures at A and B points. Density $ρ = 1000 {kg m}^{3}$ .

A and B are on the axis of the tube [2023]

1 175 Pa

2 27 Pa

3 135 Pa

4 36 Pa

Ans.
(1)

$From continuity theorem A_{1} V_{1} = A_{2} V_{2}$

$1.5 \times V_{1} = 25 \times 10^{- 2} \times 60$

$V_{1} = \frac{25 \times 60 \times 10^{- 2} \times 10}{1.5} = 10 cm/s$

$By Bernoulli's theorem$

$P_{1} + \frac{1}{2} \times 1000 \times {(0.1)}^{2} = P_{2} + \frac{1}{2} \times 1000 \times {(0.6)}^{2}$

$P_{1} + 5 = P_{2} + \frac{1}{2} \times 1000 \times 36 \times 10^{- 2}$

$\Rightarrow P_{1} - P_{2} = 175 Pa$

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